How do you solve #x(x-4)-2=43#?

2 Answers
Apr 14, 2018

Answer:

#x_"1"=-5#, #x_"2"=9#

Explanation:

#x(x-4)-2=43#
#x*x-x*4-2=43#
#x²-4x-2-43=43-43#
#x²-4x-45=0#
#Δ=(-4)²-4*1*(-45)#
#Δ=196#
#x=(-b+-sqrt(Δ))/(2a)#
#x_"1"=(-(-4)-sqrt(196))/2#
#x_"2"=(-(-4)+sqrt(196))/2#
#x_"1"=-5#, #x_"2"=9#
\0/ here's our answer!

Apr 14, 2018

Answer:

#x=#9 or -5

Explanation:

Expand the bracket

#x(x-4)-2=43# #=># #x^2-4x-2=43#

Subtract 43 from both sides

#x^2-4x-45=0#

Factorise

#(x-9)(x +5)=0#

So #x=9 or -5#